垂直多關(guān)節(jié)機器人位置控制結(jié)構(gòu)設(shè)計（腰部關(guān)節(jié)）
垂直多關(guān)節(jié)機器人位置控制結(jié)構(gòu)設(shè)計（腰部關(guān)節(jié)）,垂直,關(guān)節(jié),機器人,位置,控制,結(jié)構(gòu)設(shè)計,腰部
工業(yè)機器人手臂的靜態(tài)平衡
第一部分：平衡離散
Ion Simionescu*, Liviu Ciupitu
Mechanical Engineering Department, POLITEHNICA University of Bucharest, Splaiul Independentei 313, RO-77206,
Bucharest 6, Romania
Received 2 October 1998; accepted 19 May 1999
摘要：本文介紹了一些在工業(yè)機器人手臂的重量平衡解決方案，運用了螺旋彈簧的彈性力量。 垂直和水平手臂的重量力量的平衡顯示很多備選方案。 最后，舉例子，解決一個數(shù)值示例。
關(guān)鍵詞：工業(yè)機器人；靜態(tài)平衡；離散平衡
7 2000 Elsevier Science Ltd. All rights reserved.
1. 介紹
機器人及工業(yè)機器人機制構(gòu)成了一個特殊類別的機器系統(tǒng)，其特點是大質(zhì)量的元素在一個垂直平面移動速度相對緩慢?；谶@個原因，重量勢力成了驅(qū)動系統(tǒng)必須要克服的一大份額的阻力。對于平衡重量力量的問題，可編程序的機器人是非常重要的，在訓(xùn)練期間，人工操作必須容易地駕駛機械系統(tǒng)。
一般來說，工業(yè)機器人手臂的重量平衡力量都將會削弱驅(qū)動力量。在軸承發(fā)生的摩擦力沒有被考慮到，因為摩擦?xí)r刻感覺取決于相對運動感覺。
在這項工作中，對直圓柱螺旋彈簧彈力影響力量平衡問題的可能性進行了分析。
這種平衡的可以被分離出來，可以是工作領(lǐng)域位置的有限數(shù)字，或者在在工作領(lǐng)域中的所有位置的連續(xù)。 因此，離散系統(tǒng)只能實現(xiàn)了機器人手臂的近似平衡。
增量的使用并沒有被考慮在內(nèi)，因為他們涉及到了移動的質(zhì)量物體的增加，整體大小，慣性和組分的壓力。
2. 在一固定水平軸附近的重量力量的平衡
通過螺旋彈簧的彈力來平衡機器手和機器人的重量力量，有集中可行的方案。
簡單的解決方案并不總是適用的。有時候從建筑角度來首選一個有效的近似解替代原先方案。
在一個水平固定軸附近的鏈接1（例如：橫向機械手臂）的重量力量的維持平衡的最簡單的方法在圖1中該要的顯示出來了。在鏈接點A和固定點B之間，使用了一個螺旋彈簧2.以下是對鏈接1適用的表達力矩的平衡公式：(m1OG1cosi+m2A)g+Fsa=0,i=1,…,6
在那里，螺旋彈簧彈力是: FS=F+k(AB-l0),和
彈簧2的重心G2和雙中心A、B兩點在同一個直線上。
彈簧的彈性系數(shù)由 k 表示、 m1 是鏈接 1 的質(zhì)量、 m2 是 螺旋彈簧2的質(zhì)量 ， g 表示重力加速度的大小。這樣，通過六個非重復(fù)值Ψi以及由其獲得的力的平衡值，可以獲得以下的未知值：1A,y1A,XB,YB,F0和K 。
為了使得重心G1位于OX1 上，對于手臂1我們選擇活動協(xié)調(diào)軸系統(tǒng)X1 OY1 . X1A 和Y1A 的調(diào)整確定了臂1上點A的位置。
在一些特殊的情況下，當y1A=XB=l0=F0=0 時，這個問題可以有無限的解答，通過下面的公式定義：
k=,
角度Ψ取任意值。
因為在這種情況下， FS=k AB（見圖2 第一行），不使用螺旋彈簧的系統(tǒng)在建筑上出現(xiàn)了一些困難。壓縮彈簧，它對于計算的功能，不能被對折。因此，在導(dǎo)航中出現(xiàn)的摩擦力使得培訓(xùn)工作更加困難。
甚至于在一般的情況下，當y1A≠0和XB≠0時，彈簧的初始長度l0 的減少，相當于力F0=0。對于平衡所必須的彈簧的平直特征位置的徑向變位系數(shù)（圖2直線2），換言之，從建筑學(xué)的角度上看，為了獲得一個可以接受的原始長度l0 ,可能可以用一個移動的彈簧取代固定B點的彈簧連接。換句話來說，彈簧的B端掛在可移動的鏈接2上，位置隨著手臂1的變化而變化。鏈接2可能有一個平面副的或者是直線的繞著一個固定點的轉(zhuǎn)動運動副，并且它通過中介動力學(xué)鏈子所驅(qū)動。（圖3-5）在引用里展示了更多的可能性[2-7]。
圖3. 彈性系統(tǒng)的平衡與四桿機構(gòu)
圖3展示了一個運動學(xué)構(gòu)架，其中連接2在C點幀加入，它通過連接桿3和機器人手臂1的鏈接進行驅(qū)動。在手臂1運行的平衡力量系統(tǒng)由一下方程表示：
fi=(m1OG1cos+m4AXA)g+Fs(YAcos－XAsin)+R31XYE－R31YXE=0,
i=1,…，12， （2）
在連接桿3和機器人手臂1之間的反作用力組分，在固定坐標系軸上：
類似于前面的例子，連接桿3的角度是：
OG1 和BG4的距離，同,,,分別決定了鏈接1、4、2.2 的質(zhì)量重心的位置。
未知數(shù) ,, ,,,,,,ED, BC, 和k通過解決平衡方程（2）解得，其中需要工作區(qū)域12個機器人手臂的非重復(fù)位置角Ψi 。元素的質(zhì)量mj ( j=1,….,4)和物質(zhì)中心假設(shè)是已知的。根據(jù)那些角：i,i=1,…，12機器人手臂的靜態(tài)平衡在那些12個位置保持平衡。由于連續(xù)性的原因，不平衡值在這些位置上是微不足道的。
實際上，問題是以一種反復(fù)的方式解決的，因為在設(shè)計之初，關(guān)于螺旋彈簧和鏈接2和3的情況，很多都是未知的。
不平衡力矩的最大值和平衡系統(tǒng)的未知數(shù)成反比。通過在臂1和鏈接2上兩個平行圓柱螺旋彈簧的組裝，平衡精度增加了，因為18個非重復(fù)值的Ψi可施加在相同的工作領(lǐng)域。
在 Fig.4 中，顯示了圍繞一個固定的橫軸的鏈接的靜態(tài)平衡的另一種可能性。被固定在直線上滑行的滑道2上的B點通過機器人手臂由桿3驅(qū)動。該系統(tǒng)根據(jù)以下的平衡方程形成：
fi=(m1OG1cos+m4AXA)g+Fs(YAcos－XAsin)+R13XYE－R13YXE=0,
i=1,…，11， (3)
未知數(shù)：,,,,CD,d,b,e,a,,and k。
滑塊的位移Si可以取以下的值：
圖.5.彈性系統(tǒng)與曲柄滑塊機構(gòu)Ⅱ.
如果工作領(lǐng)域關(guān)于垂直軸OY對稱，那么平衡機制就有一個特定的模式，并由這些變量決定：y1A=y1D=b=e=0,和 [5]。
未知值減少到了六個 ，但是平衡精度提高了，因為考慮到了位置角Ψi決定了以下的方程式：
，i=1…,6. (4)
同樣，平衡螺旋彈簧4可以在B點加入到連桿點3.。(Fig.5).Eq.(3) 臂1和鏈接3之間的反應(yīng)力的構(gòu)成為：
未知數(shù)為：,,,,,CD,e,a,,and k。
圖6顯示了另一個平衡系統(tǒng)變體。螺旋彈簧4B端加入了能夠平面平行運動的連桿3.以下的未知數(shù),,,,,,,,d,,和 k.被作為由以下平衡方程構(gòu)筑的系統(tǒng)的解決方案(3)：
和
圖.6. 彈性系統(tǒng)的平衡與振蕩滑塊機構(gòu).
一樣的方法，如果工作領(lǐng)域關(guān)于垂直軸Oy對稱.(y1A=y1E=y3B=d=XC=0)[5]的話，在圖4顯示的建設(shè)性的解決方案，平衡精度性更高，因為位置角Ψi決定了方程式。
圖.7. 縱向和橫向平衡的機器人手臂彈性系統(tǒng).
3、四連桿結(jié)構(gòu)的重力的靜態(tài)平衡
由于機器人垂直壁承載著水平臂的問題，機器人垂直臂的靜態(tài)平衡顯示出了一些特殊情況?；谶@個原因，大多數(shù)的機器人制造商選擇使用平行四邊形模型作為一個垂直臂。(如圖.7)因此，鏈接3有一個圓形平移運動。在K點加入了彈性系統(tǒng)，是為了平衡水平機器手臂重量。以上的任何一個方案都可以解決四連桿元素的重量力平衡問題。例如，圖3的彈性系統(tǒng)。彈性系統(tǒng)的未知尺寸同時解決了下面的方程：
以上這個方程所寫的12個垂直臂可變位置角的值。
這些方程是虛功原理應(yīng)用于鏈接系統(tǒng)的成果。當水平的手臂不旋轉(zhuǎn)繞軸 C，而因此由 3,8,9,10 和 11 幾元素組成的重心的速度等于點 C.的速度時，等式（5）是成立的。所有的鏈接和重心的位置都應(yīng)該是已知的。等式（5）可以被等式（6）替代，如果d2/dt=1成立：
以下是未知值：
l FG和GH的長度；
l 坐標：點F,J,H 和 J的坐標；,,,,,,,
l 對應(yīng)于原始長度l0 和剛性彈簧系數(shù)k 的F0
4. 舉例
機器人手臂質(zhì)量m1=10kg 和 圖3的彈性系統(tǒng)處于靜態(tài)平衡狀態(tài)，已知：
DE =0.100706 m, BC = 0.161528 m, x1E =0.145569m, y1E =－0.84820×10－6 m, XC =0.244535×10－3 m, YC = 0.0969134 m, x1A =0.820178m, y1A= 0.144475×10－3 m, x2D=－0.0197607 m, y2D= －0.146229 m。
重心G1有OG1=1.0m 。關(guān)于彈簧有 原始長度l0 =0.5m 彈性系數(shù)k=3079.38N/m ,彈簧重m4 =1.5 kg 。
當min=－0.785398和max=－0.785396時，最大不平衡時刻有最大值，最大值UMmax=0.271177 Nm。
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The static balancing of the industrial robot armsPart I: Discrete balancingIon Simionescu*, Liviu CiupituMechanical Engineering Department, POLITEHNICA University of Bucharest, Splaiul Independentei 313, RO-77206,Bucharest 6, RomaniaReceived 2 October 1998; accepted 19 May 1999AbstractThe paper presents some new constructional solutions for the balancing of the weight forces of theindustrial robot arms, using the elastic forces of the helical springs. For the balancing of the weightforces of the vertical and horizontal arms, many alternatives are shown. Finally, the results of solving anumerical example are presented. 7 2000 Elsevier Science Ltd. All rights reserved.Keywords: Industrial robot; Static balancing; Discrete balancing1. IntroductionThe mechanisms of manipulators and industrial robots constitute a special category ofmechanical systems, characterised by big mass elements that move in a vertical plane, withrelatively slow speeds. For this reason the weight forces have a high share in the category ofresistance that the driving system must overcome. The problem of balancing the weight forcesis extremely important for the play-back programmable robots, where the human operatormust drive easily the mechanical system during the training period.Generally, the balancing of the weight forces of the industrial robot arms results in thedecrease of the driving power. The frictional forces that occur in the bearings are not takenMechanism and Machine Theory 35 (2000) 128712980094-114X/00/$ - see front matter 7 2000 Elsevier Science Ltd. All rights reserved.PII: S0094-114X(99)00067- Corresponding author.E-mail address: simionform.resist.pub.ro (I. Simionescu).into consideration because the frictional moment senses depend on the relative movementsenses.In this work, some possibilities of balancing of the weight forces by the elastic forces of thecylindrical helical springs with straight characteristics are analysed.This balancing can be made discretely, for a finite number of work field positions, or incontinuous mode for all positions throughout the work field. Consequently, the discretesystems realised only an approximatively balancing of the arm.The use of counterweights is not considered since they involve the increase of movingmasses, overall size, inertia and the stresses of the components.2. The balancing of the weight force of a rotating link around a horizontal fixed axisThere are several possibilities of balancing the weight forces of the manipulator and robotarms by means of the helical spring elastic forces.The simple solutions are not always applicable. Sometimes an approximate solution ispreferred, leading to a convenient alternative from constructional point of view.The simplest balancing possibility of the weight force of a link 1 (the horizontal robot arm,for example) which rotates around a horizontal fixed axis is schematically shown in Fig. 1. Ahelical spring 2, joined between a point A of the link and a fixed B one, is used. The equationthat expresses the equilibrium of the forces moments 1, which act to the link 1, is?m1OG1cos ji? m2AXA?g ? Fsa ? 0, i ? 1,.,6,?1?where the elastic force of the helical spring is:Fs? F0? k?AB ? l0?,andFig. 1I. Simionescu, L. Ciupitu / Mechanism and Machine Theory 35 (2000) 128712981288a ?XBYA? XAYBAB;?XAYA? Rji?x1Ay1A?;Rji?cos ji?sin jisin jicos ji?;AB ?XA? XB?2?YA? YB?2q;m2A?BG2ABm2:The gravity centre G2of spring 2 is collinear with pairs centres A and B.The sti?ness coe?cient of the spring is denoted by k, m1is the mass of the link 1, m2is themass of the helical spring 2, and g represents the gravity acceleration magnitude.Thus, the unknown factors: x1A, y1A, XB, YB, F0and k may be calculated in such a way thatthe equilibrium of the forces is obtained for six distinct values of the angle ji: The movable co-ordinate axis system x1Oy1attached to the arm 1 was chosen so that the gravity centre G1isupon the Ox1axis. The co-ordinates x1Aand y1Adefined the position of point A of the arm 1.In the particular case, characterised by y1A? XB? l0? F0? 0, the problem allows aninfinite number of solutions, which verify the equation:k ?m1OG1? m2Ax1A?gx1AYB,for any value of angle j:Since in this case, Fs? k AB (see line 1, Fig. 2), some di?culties arise in the construction ofthis system where it is not possible to use a helical extension spring. The compression spring,which has to correspond to the calculated feature, must be prevented against buckling.Consequently, the friction forces that appear in the guides make the training operation moredi?cult.Even in the general case, when y1A6?0 and XB6?0, results a reduced value of the initial lengthl0of the spring, corresponding to the forces F0? 0: The modification of the straightcharacteristic position to the necessary spring for balancing (line 2, Fig. 2), i.e. to obtain anacceptable initial length l0from the constructional point of view, may be achieved by replacingthe fixed point B of spring articulation by a movable one. In other words, the spring will bearticulated with its B end of a movable link 2, whose position depends on that of the arm 1.Link 2 may have a rotational motion around a fixed axis, a plane-parallel or a translationalone, and it is driven by means of an intermediary kinematics chain (Figs. 35).Further possibilities are shown in Refs. 27.Fig. 2I. Simionescu, L. Ciupitu / Mechanism and Machine Theory 35 (2000) 128712981289Fig. 3 shows a kinematics schema in which link 2 is joined with the frame at point C, and itis driven by means of the connecting rod 3 from the robot arm 1. The balancing of the forcessystem that acts on the arm 1 is expressed by the following equation:fi?m1OG1cos ji? m4AXA?g ? Fs?YAcos yi? XAsin yi? ? R31XYE? R31YXE? 0,i ? 1,.,12,?2?where: yi? arctanYB?YAXB?XA; m4A?BG4ABm4; m4B? m4? m4A;?XEYE? Rji?x1Ey1E?;?XBYB?XCYC? Rci?BC0?;Rci?cos ci?sin cisin cicos ci?:The components of the reaction force between the connecting rod 3 and the arm 1, on the axesof fixed co-ordinate system, are:R31X?T?XD? XE? ? m3?XD? XG3?XC? XE?gYD?XC? XE? ? YC?XD? XE? ? YE?XC? XD?;R31Y?R31X?YE? YD? ? m3?XG3? XD?gXD? XE,where:T ? Fs?XB? XC?sin yi? ?YB? YC?cos yi?hm2?XG2? XC? m3?XG3? XC? m4B?XB? XC?ig,Fig. 3. Balancing elastic system with four bar mechanism.I. Simionescu, L. Ciupitu / Mechanism and Machine Theory 35 (2000) 128712981290?XDYD?XCYC? Rci?x2Dy2D?;?XG2YG2?XCYC? Rci?x2G2y2G2?;?XG3YG3?XCYC? Rxi?x3G3y3G3?,Rxi?cos xi?sin xisin xicos xi?:The value of angle ci:ci? arctanU?U2? V2? W2p? VW?V?U2? V2? W2p? UW? arepresents the solution of the equation:U cos?ci? a? V sin?ci? a? W ? 0,where:U ? 2CD?XC? XE?;V ? 2CD?YE? YC?;W ? OE2? CD2? OC2? DE2? 2?XEXC? YEYC?;a ? arctany2Dx2D:Similar to the previous case, the angle of the connecting rod 3 is:xi? arccosCD cos?ci? a? XC? XEDEThe distances OG1and BG4, and the co-ordinates: x2G2, y2G2, XG3, YG3give the positions of themass centres of links 1, 4, 3 and 2, respectively.The unknowns of the problem: x1A, y1A, x1E, y1E, x2D, y2D, XC, YC, ED, BC, F0and k arefound by solving the system made up through reiterated writing of the equilibrium equation (2)for 12 distinct values of the position angle jiof the robot arm 1, which are contained in thework field. The masses mj, j ? 1,.,4, of the elements and the positions of the mass centresare assumed as known. The static equilibrium of the robot arm is accurately realised in those12 positions according to angles ji, i ? 1,.,12 only. Due to continuity reasons, theunbalancing value is negligible between these positions.In fact, the problem is solved in an iterative manner, because at the beginning of the design,the masses of the helical spring and links 2 and 3 are unknown.The maximum magnitude of the unbalanced moment is inverse proportional to the numberof unknowns of the balancing system. By assembling the two helical springs in parallel betweenI. Simionescu, L. Ciupitu / Mechanism and Machine Theory 35 (2000) 128712981291arm 1 and link 2, the balancing accuracy is increased, since 18 distinct values of angle jimaybe imposed within the same work field.In Fig. 4, another possibility for the static balancing of a link that rotates around ahorizontal fixed axis is shown. The point B belongs to slide 2 which slides along a fixedstraight line and is driven by means of the connecting rod 3 by the robot arm 1. The system,formed by following equilibrium equations:fi?m1OG1cos ji? m4AXA?g ? Fs?YAcos y ? XAsin y? ? R13XYE? R13YXE? 0,i ? 1,.,11,?3?whereR13X?m2? m3? m4B?g sin a ? Fscos?y ? a?DE ? m3gDG3sin aDE cos?a ? ci?cos ci;R13Y?m3gDG3cos a cos ci?m2? m3? m4B?g sin a ? Fscos?y ? a?DE sin ciDE cos?a ? ci?;ci? a ? arcsinXEsin a ? YEcos a ? b ? eDE;XB? e sin a ? ?Si? d?cos a;YB? ?Si? d?sin a ? e cos a,are solved with respect to the unknowns: x1A, y1A, x1D, y1D, CD, d, b, e, a, F0and k.The displacement Siof the slider has the value:Fig. 4. Elastic system with slider-crank mechanism I.I. Simionescu, L. Ciupitu / Mechanism and Machine Theory 35 (2000) 128712981292Si?XE? DE cos ci? ?b ? e?sin acos a,if a6?p2,orSi?YE? DE sin ci? ?b ? e?cos asin a,if a6?0:If the work field is symmetrical with respect to the vertical axis OY, the balancingmechanism has a particular shape, characterised by y1A? y1D? b ? e ? 0, and a ? p=2 5.The number of the unknowns decreased to six, but the balancing accuracy is higher, becauseit is possible to consider that the position angles jiverify the equality:ji?6? p ? ji,i ? 1,.,6:?4?Likewise, the balancing helical spring 4 can be joined to the connecting rod 3 at point B(Fig. 5). Eq. (3) where the components of the reaction force between the arm 1 and link 3 are:R13X?m2? m3? m4B?g sin a ? Fscos?y ? a?cos cicos?a ? ci?m3?XG3? XD? ? m4B?XB? XD?g ? Fs?XB? XD?sin y ? ?YB? YD?cos y?DE cos?a ? ci?sin a;Fig. 5. Elastic system with slider-crank mechanism II.I. Simionescu, L. Ciupitu / Mechanism and Machine Theory 35 (2000) 128712981293R13Y?m2? m3? m4B?g sin a ? Fscos?y ? a?sin cicos?a ? ci?m3?XG3? XD? ? m4B?XB? XD?g ? Fs?XB? XD?sin y ? ?YB? YD?cos y?DE cos?a ? ci?cos a;?XBYB?XDYD? Rci?x3By3B?,ci? a ? arcsinXEsin a ? YEcos a ? eDE,is solved with respect to the unknowns: x1A, y1A, x1D, y1D, x3B, y3B, CD, e, a, F0and k.Fig. 6 shows another variant for the balancing system. The B end of the helical spring 4 isjoined to the connecting rod 3 which has a plane-parallel movement. The following unknowns:x1A, y1A, x1E, y1E, x3B, y3B, XC, YC, d, F0and k are found as solutions of the system made upof equilibrium equation (3), where:R13X?U sin ci? V?XE? XC?W;R13Y?V?YC? YE? ? U cos ciW;and:U ? Fs?XB? XC?sin y ? ?YB? YC?cos y?hm2?XG2? XC? m3?XG3? XC? m4B?XB? XC?ig;V ? Fscos?ci? y? m3g sin ci;Fig. 6. Balancing elastic system with oscillating-slider mechanism.I. Simionescu, L. Ciupitu / Mechanism and Machine Theory 35 (2000) 128712981294W ? ?YC? YE?sin ci? ?XC? XE?cos ci;ci? arctanYC? YEXC? XE? arcsindCE;CE ?XC? XE?2?YC? YE?2q:In the same manner as the constructive solution shown in Fig. 4, the balancing accuracy ishigher, if the work field is symmetrical with respect to the vertical OY axis ?y1A? y1E? y3B?d ? XC? 0? 5, because the position angles jiverify the equality (4).Fig. 7. Balancing elastic systems for vertical and horizontal robot arms.I. Simionescu, L. Ciupitu / Mechanism and Machine Theory 35 (2000) 1287129812953. The static balancing of the weight forces of four bar linkage elementsThe static balancing of a vertical arm of a robot presents some particularities, consideringthat it bears the horizontal arm. For this reason, most of the robot manufacturers use aparallelogram mechanism as a vertical arm (Fig. 7). Therefore, the link 3 has a circulartranslational movement. At point K is joined the elastic system that is used for balancing theweight of the horizontal robot arm. For balancing of the weight forces of the four-bar linkageelements, any one of the constructive solutions mentioned above can be used. For example, theelastic system schematised in Fig. 3 is considered. The unknown dimensions of the elasticsystem are found by simultaneously solving the following equations:?m2dYG2dt? ?m3? m8? m9? m10? m11?dYCdt? m4dYG4dt? m5dYG5dt? m6dYG6dt?m72?dYIdt?dYJdt?g ? FsdIJdt? 0,?5?which are written for 12 distinct values of the position angle j2iof the vertical arm.These equations result from applying on the virtual power principle to force system whichacts on the linkage. The equality (5) is valid when the horizontal arm does not rotate aroundthe axis of pair C, and consequently the velocity of the gravity centre of the ensemble formedby the elements 3, 8, 9, 10 and 11 is equal to the velocity of point C. The masses of the linksand the positions of the gravity centres are supposed to be known.Eq. (5) may be substituted by Eq. (6), if it is assumed that dj2=dt ? 1:?m2dYG2dj2? ?m3? m8? m9? m10? m11?dYCdj2? m4dYG4dj2? m5dYG5dj2? m6dYG6dj2?m72?dYIdj2?dYJdj2?g ? FsdIJdj2? 0,?6?where:Fs? F0?XI? XJ?2?YI? YJ?2q? l0?k;YG2? x2G2sin j2i? y2G2cos j2i;YG4? x4G4sin j2i? y4G4cos j2i;YG5? YF? x5G5sin j5i? y5G5cos j5i;I. Simionescu, L. Ciupitu / Mechanism and Machine Theory 35 (2000) 128712981296YG6? YH? x6G6sin j6i? y6G6cos j6i;YI? YH? x6Isin j6i? y6Icos j6i;YJ? x2Jsin j2i? y2Jcos j2i;XF? x2Fcos j2i? y2Fsin j2i;YF? x2Fsin j2i? y2Fcos j2i;YC? BC sin j2i;j5i? arctanVW ? U?U2? V2? W2pUW ? V?U2? V2? W2p;U ? 2FG?XF? XH?;V ? 2FG?YF? YH?;W ? GH2? FG2? ?XF? XH?2?YF? YH?2;j6i? arctanST ? R?R2? S2? T2pRT ? S?R2? S2? T2p;R ? 2GH?XH? XF?;S ? 2GH?YH? YF?;T ? FG2? GH2? ?XF? XH?2?YF? YH?2:The unknowns of the problem are:. the lengths FG and GH;. the co-ordinates: x2F, y2F, x2J, y2J, XH, YH, x6I, y6Iof the points F, J, H and J, respectively;. the force F0, corresponding to the initial length l0, and the sti?ness coe?cient k of the helicalspring 7.4. ExampleA robot arm of mass m1? 10 kg is statically balanced with the elastic system schematised inFig. 3, having the following dimensions: DE ? 0:100706 m, BC = 0.161528 m, x1E? 0:145569m, y1E? ?0:848205 ? 10?6m, XC? 0:244535 ? 10?3m, YC? 0:0969134 m, x1A? 0:820178m, y1A? 0:144475 ? 10?3m, x2D? 0:0197607 m, y2D? ?0:146229 m. The distance to theI. Simionescu, L. Ciupitu / Mechanism and Machine Theory 35 (2000) 128712981297gravity centre G1is OG1? 1:0 m. The characteristics of the spring are: the initial length l0?0:5 m, the sti?ness coe?cient k ? 3079:38 N/m, and the mass m4? 1:5 kg.In the work field defined by jmin? ?0:785398 and jmax? 0:785396, the maximumunbalanced moment has the magnitude UMmax? 0:271177 Nm.References1 P. Appell, Traite de me canique rationnelle, Gauthier Villars, Paris, 1928.2 A. Gopaswamy, P. Gupta, M. Vidyasagar, A new parallelogram linkage configuration for gravity compensationusing torsional springs, in: Proceedings of IEEE International Conference on Robotics and Automation, vol. 1,Nice, France, 1992, pp. 664669.3 K. Hain, Spring mechanisms point balancing, in: N.D. Chironis (Ed.), Spring Design and Application,McGraw-Hill, New York, 1961, pp. 268275.4 E.P. Popov, A.N. Korenbiashev, Robot Systems, Mashinostroienie, Moscow, 1989.5 I. Simionescu, L. Ciupitu, On the static balancing of the industrial robots, in: Proceeding of the 4thInternational Workshop on Robotics in AlpeAdria Region RAA 95, July 68, Po rtschach, Austria, vol. II,1995, pp. 217220.6 I. Simionescu, L. Ciupitu, The static balancing of the industrial robot arms, in: Ninth World Congress on theTheory of Machines and Mechanisms, Aug. 29Sept. 2, Milan, Italy, vol. 3, 1995, pp. 17041707.7 D.A. Streit, E. Shin, Journal of Mechanical Design 115 (1993) 604611.I. Simionescu, L. Ciupitu / Mechanism and Machine Theory 35 (2000) 128712981298
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